 Solution

For the first matrix, the characteristic polynomial isThus the real eigenvalues are $4,-1,3$. Corresponding eigenvectors are $(3,1,0{)}^{\mathsf{T}}$, $(-2,1,0{)}^{\mathsf{T}}$ and $(0,0,1{)}^{\mathsf{T}}$. They are linearly independent; the determinant of the matrix with these columns is $5$. ThereforeFor example, multiplying the first column by $A$ gives $(12,4,0{)}^{\mathsf{T}}$, four times that column; multiplying the second gives $(2,-1,0{)}^{\mathsf{T}}$, its negative. These also check the chosen eigenvectors directly.

For the second matrix,The quadratic factor has negative discriminant $-3$, so $1$ is the only real eigenvalue. Solving $Av=v$ gives $v=(a,a,a{)}^{\mathsf{T}}$, a one-dimensional eigenspace. It cannot provide a basis of ${\mathbb{R}}^3$, so